Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using ll = long long;

int n;
int a[212345];

int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  cin >> n;
  for(int i = 0; i < n; i++) cin >> a[i];
  vector<int> ord(n);
  iota((ord).begin(), (ord).end(), 0);
  sort((ord).begin(), (ord).end(), [&](int x, int y) { return a[x] > a[y]; });
  int T = 0, f = 0;
  int last = ord[0];
  for(int i = 1; i < n; i++) {
    if(T > 0) {
      // ord[i] , ord[i + 1], f
      vector<int> v{ord[i], ord[i - 1], f};
      for(int t = 0; t < 3; t++) {
        tie(v[0], v[1], v[2]) = make_tuple(v[1], v[2], v[0]);
        if(v[0] < v[1] && v[1] < v[2]) break;
        if(v[0] > v[1] && v[1] > v[2]) {
          f = ord[i - 1];
          T++;
          break;
        }
      }
    } else {
      if(ord[i] < last);
      else T = 1, f = last;
      last = ord[i];
    }
  }
  cout << T << endl;
  return 0;
}